# How to simplify (2x - 3)^2 - 2x(2x - 5)

tehmatriks

## Homework Statement

simplify (2x-3)² - 2x(2x-5).

## The Attempt at a Solution

(2x-3)² - 2x(2x-5)
to
2x² + 9 - 4x² + 10x
to
2x² + 9 + 10x

## Answers and Replies

Homework Helper
hi tehmatriks! sorry, but your (2x-3)² is completely wrong if you can't do it in your head, write it as (2x-3)(2x-3) first, and then expand it tehmatriks

hi tehmatriks! sorry, but your (2x-3)² is completely wrong if you can't do it in your head, write it as (2x-3)(2x-3) first, and then expand it thanks, the whole time i was staring at the (2x-3)² and was just thinking about how wrong i was doing it, i always forget that double bracket thing when it comes to these situations, just dont get these much

thanks again man, here's the final result

(2x-3)² - 2x(2x-5)
to
(2x-3)(2x-3) - 2x(2x-5)
to
4x - 6x - 6x + 9 - 4x + 10x
to
9 - 2x

Mentor

thanks, the whole time i was staring at the (2x-3)² and was just thinking about how wrong i was doing it, i always forget that double bracket thing when it comes to these situations, just dont get these much

thanks again man, here's the final result

(2x-3)² - 2x(2x-5)
to
(2x-3)(2x-3) - 2x(2x-5)
to
4x - 6x - 6x + 9 - 4x + 10x
to
9 - 2x

Use = between expressions that have the same value.

(2x-3)2 - 2x(2x-5)
= (2x-3)(2x-3) - 2x(2x-5) -- so far, so good
After that, things go downhill.
(2x)(2x) = 2*2*x*x = ?

And -2x(2x - 5) = -2x * 2x -2x * (-5) = ?
You have to distribute the -2x over both terms inside the parentheses.

tehmatriks

Use = between expressions that have the same value.

(2x-3)2 - 2x(2x-5)
= (2x-3)(2x-3) - 2x(2x-5) -- so far, so good
After that, things go downhill.
(2x)(2x) = 2*2*x*x = ?

And -2x(2x - 5) = -2x * 2x -2x * (-5) = ?
You have to distribute the -2x over both terms inside the parentheses.

dammit, it's always the simplest of things that i forget

(2x-3)(2x-3) - 2x(2x-5)
to
4x2 - 6x - 6x + 9 - 4x2 + 10x
to -12x + 10x + 9 = 9 - 2x

thanks mark, you're a legend Mentor

Much better.

The things that you're forgetting are the result of not learning them very well in the first place. If you want to get better at this, I would advise going back and reviewing binomial multiplication (problems like expanding (2x + 3)2) and the distributive law (a(b + c) = ab + ac).